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Electrostatics NEET Questions — Free MCQ Practice

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An electric line of force in the xy plane is given by equation x² + y² = 1. A particle with unit positive charge, initially at rest at the point x = 1, y = 0 in the xy plane will -

A.

Not move at all

B.

Will move along straight line

C.

Will move along the circular line of force

D.

Information is insufficient to draw any conclusion

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Explanation

Charge will move along the circular line of force because x² + y² = 1 is the equation of circle in xy-plane.

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A positively charged ball hangs from a silk thread. We put a positive test charge q₀ at a point and measure F/q₀, then it can be predicted that the electric field strength E

A.

> F/q₀

B.

= F/q₀

C.

< F/q₀

D.

Cannot be estimated

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Explanation

Because of the presence of positive test charge q₀ in front of positively charged ball, charge on the ball will be redistributed, less charge on the front half surface and more charge on the back half surface. As a result of this net force F between ball and point charge will decrease i.e. actual electric field will be greater than F/q₀.

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A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shell having charge –Q. The radius of the sphere is a and that of the spherical shell is b (b > a). What is the electric field at a distance R(a < R < b) from the centre

A.

$\frac{Q}{2 π ε_{0} R}$

B.

$\frac{3 Q}{2 π ε_{0} R}$

C.

$\frac{3 Q}{4 π ε_{0} R^{2}}$

D.

$\frac{4 Q}{4 π ε_{0} R^{2}}$

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Explanation

Electric field at a distance R is only due to sphere because electric field due to shell inside it is always zero. Hence electric field = $\frac{1}{4 π ε_{0}} . \frac{3 Q}{R^{2}}$

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A point charge q is placed at a distance a/2 directly above the centre of a square of side a. The electric flux through the square is

A.

$\frac{q}{ε_{0}}$

B.

$\frac{q}{π ε_{0}}$

C.

$\frac{q}{4 ε_{0}}$

D.

$\frac{q}{6 ε_{0}}$

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Two infinitely long parallel wires having linear charge densities λ₁ and λ₂ respectively are placed at a distance of R meters. The force per unit length on either wire will be $(K = \frac{1}{4 π ε_{0}})$

A.

$K \frac{2 λ_{1} λ_{2}}{R^{2}}$

B.

$K \frac{2 λ_{1} λ_{2}}{R}$

C.

$K \frac{λ_{1} λ_{2}}{R^{2}}$

D.

$K \frac{λ_{1} λ_{2}}{R}$

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Explanation

When two infinitely long parallel wires with linear charge densities Î»1 and Î»2 are placed at a distance R, the force per unit length on either wire is given by the expression (2kÎ»1Î»2)/R, where k = 1/(4Ï€Îµ0) is the Coulomb constant. This formula arises from the application of Coulomb's law and the principle of linear superposition.

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The electric field in a region is radially outward with magnitude $E = A γ_{0}$ . The charge contained in a sphere of radius $γ_{0}$ centered at the origin is

A.

$\frac{1}{4 π ε_{0}} A γ_{0}^{3}$

B.

$4 π ε_{0} A γ_{0}^{3}$

C.

$\frac{4 π ε_{0} A}{γ_{0}}$

D.

$\frac{1}{4 π ε_{0}} \frac{A}{γ_{0}^{3}}$

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Explanation

Flux linked with the given sphere $φ = \frac{Q}{ε_{o}};$

where Q = Charge enclosed by the sphere.

Hence Q = φε₀ = (EA)ε₀

⇒ Q = 4π (γ₀)² × Aγ₀ε₀ = 4πε₀Aγ₀³.

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Charge q is uniformly distributed over a thin half-ring of radius R. The electric field at the centre of the ring is

A.

$\frac{q}{2 π^{2} ε_{0} R^{2}}$

B.

$\frac{q}{4 π^{2} ε_{0} R^{2}}$

C.

$\frac{q}{4 π ε_{0} R^{2}}$

D.

$\frac{q}{2 π ε_{0} R^{2}}$

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Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when

A.

$x = \frac{d}{\sqrt{2}}$

B.

$x = \frac{d}{2}$

C.

$x = \frac{d}{2 \sqrt{2}}$

D.

$x = \frac{d}{2 \sqrt{3}}$

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Explanation

When two equal charges are separated by a distance d, the maximum Coulomb force experienced by a third charge placed on the perpendicular bisector occurs when the third charge is at a distance x = d/(2âˆš2) from the midpoint of the line joining the two charges. This position maximizes the net force due to the two equal charges.

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An electric dipole is situated in an electric field of uniform intensity E whose dipole moment is p and moment of inertia is I. If the dipole is displaced slightly from the equilibrium position, then the angular frequency of its oscillations is

A.

${(\frac{p E}{I})}^{1 / 2}$

B.

${(\frac{p E}{I})}^{3 / 2}$

C.

${(\frac{I}{p E})}^{1 / 2}$

D.

${(\frac{p}{I E})}^{1 / 2}$

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Explanation

When dipole is given a small angular displacement θ about it's equilibrium position, the restoring torque will be

$τ = - p E \sin θ = - p E θ$ (as sinθ = θ)

or $I \frac{d^{2} θ}{d t^{2}} = - p E θ$ (as $τ = I α = I \frac{d^{2} θ}{d t^{2}}$ )

or $\frac{d^{2} θ}{d t^{2}} = - ω^{2} θ$ with $ω^{2} = \frac{p E}{I}$ ⇒ $ω = \sqrt{\frac{p E}{I}}$

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An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along x-axis at x = 1 cm, x = 2 cm, x = 4 cm, x = 8 cm ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at x = 0 is $(\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N - m^{2} / c^{2})$

A.

12 × 10⁴

B.

24 × 10⁴

C.

36 × 10⁴

D.

48 × 10⁴

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Explanation

$E = \frac{1}{4 π ε_{0}} . [\frac{5 \times 10^{- 9}}{{(1 \times 10^{- 2})}^{2}} - \frac{5 \times 10^{- 9}}{{(2 \times 10^{- 2})}^{2}} + \frac{5 \times 10^{- 9}}{{(4 \times 10^{- 2})}^{2}} - \frac{(5 \times 10^{- 9})}{{(8 \times 10^{- 2})}^{2}} + . ....]$

$\Rightarrow E = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{10^{- 4}} [1 - \frac{1}{{(2)}^{2}} + \frac{1}{{(4)}^{2}} - \frac{1}{{(8)}^{2}} + . ..]$

$\Rightarrow E = 45 \times 10^{4} [1 + \frac{1}{{(4)}^{2}} + \frac{1}{{(16)}^{2}} + . ..] - 45 \times 10^{4} [\frac{1}{{(2)}^{2}} + \frac{1}{{(8)}^{2}} + \frac{1}{{(32)}^{2}} + . ..]$

$\Rightarrow E = 45 \times 10^{4} [\frac{1}{1 - \frac{1}{16}}] - \frac{45 \times 10^{4}}{(2)^{2}} [1 + \frac{1}{4^{2}} + \frac{1}{{(16)}^{2}} + . . .]$

E = 48 × 10⁴ – 12 × 10⁴ = 36 × 10⁴ N/C

Electrostatics MCQs for NEET — Physics Questions with Answers

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